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hitman6003

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Everything posted by hitman6003

  1. [!--quoteo(post=346608:date=Feb 16 2006, 08:24 PM:name=lightnb)--][div class=\'quotetop\']QUOTE(lightnb @ Feb 16 2006, 08:24 PM) [snapback]346608[/snapback][/div][div class=\'quotemain\'][!--quotec--] Is there an application that does this? [/quote] What your asking for is pretty specific, so, probably not. [!--quoteo(post=346608:date=Feb 16 2006, 08:24 PM:name=lightnb)--][div class=\'quotetop\']QUOTE(lightnb @ Feb 16 2006, 08:24 PM) [snapback]346608[/snapback][/div][div class=\'quotemain\'][!--quotec--] How hard would this be for me to write? [/quote] Not very. You have some options for showing the specific input fields based on which type of item is selected... You could use javascript and write all of the form fields out, then just change the content of a div based on the selected item. Or, if you are concerned about your users having to download a massive form (cause all of them would have to be on that page at the same time, just hidden, execpt for the one they selected), you could use ajax to go up to the server and retrieve the inputs. Or, you could make it a two step process. Step one (page one), get the item type. Step two (page two), based on the item type, display a certain page. The last one would be easiest. All it takes is a little html and very little php. The second is probably hardest because of the ajax and most people's fear of it. No, I won't write it for you, but if you need help with your code, I'll offer advice for what you post in the forum.
  2. When the session is created: $_SESSION['attampts'] = 0; Then on the login page: $_SESSION['attampts']++; if ($_SESSION['attampts'] > 3) { echo "You have attempted to login too many times"; } else { ...normal code... }
  3. it's easiest if I just put up an example: SELECT * FROM db LEFT JOIN db2 ON db.field1 = db2.id That works fine, but I have another field in "db" that I need to join to "db2". I've tried: SELECT * FROM db LEFT JOIN db2 ON db.field1 = db2.id LEFT JOIN db2 ON db.field2 = db2.id And: SELECT * FROM db LEFT JOIN db2 ON db.field1 = db2.id, db.field2 = db2.id And: SELECT * FROM db LEFT JOIN db2 ON db.field1, db.field2 = db2.id But I get an error with each one. Thanks for the help.
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